A · B = |A| |B| cos(Θ)
A x B = |A| |B| sin(Θ)
If [ A · B = A x B ] then cos(Θ) = sin(Θ).
Θ = 45°
If x is the angle between the two vectors then the magnitudes are equal if cos(x) = sin(x). That is, when x = pi/4 radians.
Dot Product:Given two vectors, a and b, their dot product, represented as a ● b, is equal to their magnitudes multiplied by the cosine of the angle between them, θ, and is a scalar value.a ● b = ║a║║b║cos(θ)Cross Product:Given two vectors, a and b, their cross product, which is a vector, is represented as a X b and is equal to their magnitudes multiplied by the sine of the angle between them, θ, and then multiplied by a unit vector, n, which points perpendicularly away, via the right-hand rule, from the plane that a and b define.a X b = ║a║║b║sin(θ)n
Dot Product:Given two vectors, a and b, their dot product, represented as a ● b, is equal to their magnitudes multiplied by the cosine of the angle between them, θ, and is a scalar value.a ● b = ║a║║b║cos(θ)Cross Product:Given two vectors, a and b, their cross product, which is a vector, is represented as a X b and is equal to their magnitudes multiplied by the sine of the angle between them, θ, and then multiplied by a unit vector, n, which points perpendicularly away, via the right-hand rule, from the plane that a and bdefine.a X b= ║a║║b║sin(θ)n
(A1) The dot product of two vectors is a scalar and the cross product is a vector? ================================== (A2) The cross product of two vectors, A and B, would be [a*b*sin(alpha)]C, where a = |A|; b = |B|; c = |C|; and C is vector that is orthogonal to A and B and oriented according to the right-hand rule (see the related link). The dot product of the two vectors, A and B, would be [a*b*cos(alpha)]. For [a*b*sin(alpha)]C to equal to [a*b*cos(alpha)], we have to have a trivial solution -- alpha = 0 and either a or b be zero, so that both expressions are zeroes but equal. ================================== Of course one is the number zero( scalar), and one is the zero vector. It is a small difference but worth mentioning. That is is to say if a or b is the zero vector, then a dot b must equal zero as a scalar. And similarly the cross product of any vector and the zero vector is the zero vector. (A3) The magnitude of the dot product is equal to the magnitude of the cross product when the angle between the vectors is 45 degrees.
No. A cross product is just a way of simplifying a proportion. If the cross product aren't equal, it follows logically that the proportion isn't equal.
It depends on what the dot product is meant to be equal to.
Angle of repose is equal to angle of friction.
if any one of the vectors is a null vector or if A is the angle between the two vectors then tanA =1
no .....the scalar product of two vectors never be negative Yes it can If A is a vector, and B = -A, then A.B = -A2 which is negative. Always negative when the angle is between the vectors is obtuse.
For sinX you set it equal to the opposite side of the angle over the hypotenuse(SOH),cross multiply. CosX you set it equal to the adjacent side of the angle over the hypotenuse (CAH), cross multiply. Lastly for TanX set it equal to the opposite of the angle over the adjacent side of the angle and then cross multiply (TOA). I hope that's helpful :)
if you add the vectors magnitude and equal to resultant the angle between them is 0
Both angles are equal.